attempt:
from defn of uniformally continuous $\forall \epsilon > 0 $ $\exists \delta > 0 $ s.t. $|x-y| < \delta $ with $x,y \in (-\infty,0)$ $\implies |f(x) - f(y)| < \epsilon$. My idea was to choose $x = {x_n}, y = y_m$ so that $x_n,y_m < 0$ and $\lim_{n\to \infty}x_n = \lim_{m\to \infty} y_m = 0$ i.e. for some $m,n > K \in \mathbb{N}$ we have $|x_n - y_m| < \delta$ which implies (as $f$ is uniformally continuous) $|f(x_n) - f(y_m) |< \epsilon$ so $\{f(x_n)\}$ is Cauchy, therefore converges by Cauchy's Principle.
I'm not really sure how to formalise this idea, or if it's even correct, any help please!
First uniform continuity tells you taking $\epsilon>0 \ \exists \delta>0$:
$$|x-y| < \delta \ x,y \in (-\infty,0) \Rightarrow |f(x) - f(y)| < \epsilon$$
Second take your cauchy sequence $x_n$ convergent to $0$. How is Cauchy: $$\forall \varepsilon>0 \ \exists M>0 \ \forall n,m\geq M: \ \ |x_n-x_m|<\epsilon$$
In the definition of cauchy take $\varepsilon=\delta$ then exists $M>0$: $$|x_n-x_m|<\varepsilon=\delta \ \ \forall n,m\geq M \Rightarrow |f(x_n) - f(x_m)| < \epsilon \ \forall n,m\geq M $$
Then $f(x_n)$ is Cauchy and therefore converges.